GATE CSE
First time here? Checkout the FAQ!
x
+1 vote
65 views

Let $\lambda_1, \lambda_2, \lambda_3$ denote the eigenvalues of the matrix $ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos  t & \sin  t \\ 0 & -\sin  t & \cos  t \end{pmatrix}$.

If $\lambda_1+ \lambda_2+\lambda_3=\sqrt{2} +1$ then the set of possible values of $t, - \pi \leq t < \pi$, is

  1. Empty set
  2. $\{ \frac{\pi}{4} \}$
  3. $\{ - \frac{\pi}{4}, \frac{\pi}{4} \}$
  4. $\{ - \frac{\pi}{3}, \frac{\pi}{3} \}$

 

asked in Linear Algebra by Loyal (3.6k points)  
edited by | 65 views

1 Answer

+1 vote

Answer C)

 

1-⋋  0        0

0    cos t-⋋ sin t

0     -sin t    cos t - ⋋

 

 

solving (1-⋋) {(cos t - ⋋)2 - sin2⋋}=0................i

⋋=1

⋋1+⋋2+⋋3=1+√2

if ⋋1=1

⋋2+⋋3=√2

 

from equation i we can say (cos t - ⋋)2 - sin2⋋=0

(cos t - ⋋) =(+/-) sin⋋

⋋= cos t -sin t

t=(⊼/2⨉2 + ⊼/4)  as range of t is -⊼<t<⊼

So, ans should be C)

answered by Veteran (51.7k points)  
Top Users Feb 2017
  1. Arjun

    5224 Points

  2. Bikram

    4230 Points

  3. Habibkhan

    3748 Points

  4. Aboveallplayer

    2986 Points

  5. Debashish Deka

    2356 Points

  6. sriv_shubham

    2298 Points

  7. Smriti012

    2142 Points

  8. Arnabi

    2008 Points

  9. sh!va

    1654 Points

  10. mcjoshi

    1628 Points

Monthly Topper: Rs. 500 gift card

20,832 questions
25,989 answers
59,623 comments
22,046 users